A student on a piano stool rotates freely with an angular speed of 3.07 rev/s. The student holds a 1.38 kg mass in each outstretched arm, 0.759 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 4.72 kg·m2, a value that remains constant.
a. As the student pulls his arms inward, his angular speed increases to 3.54 rev/s. How far are the masses from the axis of rotation at this time, considering the masses to be points?
b. Calculate the initial and final kinetic energy of the system.
To solve this problem, we can use the principle of conservation of angular momentum, which states that the total angular momentum of a system remains constant if no external torques act on it.
a. Let's first find the initial angular momentum of the system before the student pulls his arms inward. The initial angular momentum (L_initial) can be calculated as the product of the initial moment of inertia (I_initial) and the initial angular speed (ω_initial):
L_initial = I_initial * ω_initial
Given that I_initial is 4.72 kg·m^2 and ω_initial is 3.07 rev/s, we can substitute these values:
L_initial = 4.72 kg·m^2 * 3.07 rev/s
Now, as the student pulls his arms inward, the moment of inertia of the system decreases. The final angular momentum (L_final) is equal to the initial angular momentum since no external torques act on the system:
L_initial = L_final
We can express the final angular momentum in terms of the final moment of inertia (I_final) and the final angular speed (ω_final):
L_final = I_final * ω_final
We are given ω_final, which is 3.54 rev/s, and we need to find I_final. Let's rearrange the equation to solve for I_final:
I_final = L_final / ω_final
Substituting the values, we have:
I_final = L_initial / ω_final
Now, let's calculate I_final:
I_final = (4.72 kg·m^2 * 3.07 rev/s) / 3.54 rev/s
Simplifying, we get:
I_final = 4.08 kg·m^2
Since the masses are considered as points, we can apply the parallel axis theorem to calculate the distance (r) of the masses from the axis of rotation. The parallel axis theorem states that the moment of inertia of a compound object rotating about an axis parallel to and a distance (d) away from an axis through its center of mass is given by:
I = I_cm + M * d^2
Where I_cm is the moment of inertia about the center of mass, M is the mass, and d is the distance.
We can set up the equation for both arms of the student. Let's solve for the distance (r):
I_final = I_cm + m * r^2
Rearranging, we have:
r^2 = (I_final - I_cm) / m
Substituting the given values, we get:
r^2 = (4.08 kg·m^2 - 4.72 kg·m^2) / (1.38 kg)
Simplifying, we have:
r^2 = -0.46 kg·m^2 / (1.38 kg)
r^2 = -0.33 m^2
Since distance cannot be negative, we discard the negative sign, and the distance is:
r = √0.33 m
Therefore, the masses are approximately 0.574 m from the axis of rotation at this time.
b. To calculate the initial and final kinetic energy of the system, we can use the formula:
Kinetic Energy = (1/2) * I * ω^2
For the initial kinetic energy (KE_initial), we substitute the initial values of moment of inertia and angular speed:
KE_initial = (1/2) * (4.72 kg·m^2) * (3.07 rev/s)^2
Calculating this expression, we find:
KE_initial ≈ 23.18 J
For the final kinetic energy (KE_final), we substitute the values of I_final and ω_final:
KE_final = (1/2) * (4.08 kg·m^2) * (3.54 rev/s)^2
Calculating this expression, we find:
KE_final ≈ 26.60 J
Therefore, the initial kinetic energy is approximately 23.18 J, and the final kinetic energy is approximately 26.60 J.